The scores on a statewide language exam were normally distributed with $\mu = 83.46$ and $\sigma = 6$. Kevin earned a $90$ on the exam. Kevin's exam grade was higher than what fraction of test-takers? Use the cumulative z-table provided below. z.00.01.02.03.04.05.06.07.08.09 1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706 1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767
Solution: A cumulative z-table shows the probability that a standard normal variable will be less than a certain value (z) In order to use the z-table, we first need to determine the z-score of Kevin's exam grade. Recall that we can calculate his z-score by subtracting the mean $(\mu)$ from his grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}} = \dfrac{90 - {83.46}}{{6}} = 1.09} $ Look up $1.09$ on the z-table. This value, $0.8621$ , represents the portion of the population that scored lower than $90$ on the exam. Kevin scored higher than $86.21\%$ of the test-takers on the language exam.